append' produces '()', not the expected contextual result type 'String

Question

I was making an app that would encode messages using regular ciphers that could be decoded on paper, and I ran into this problem. Picture of error

I'm not sure if you need it, but here is the code it is inside:

func encodeMessage(input:String) -> String {
    var output:String
    if  cipher == "Ceaser/Shift" {
        for i in 0..<input.characters.count {
            output = output.append(alphabet[seekAlphabet(letter: input[i])])
        }
    return output
    }
}

Another piece of code that might help:

func seekAlphabet(letter:String) -> Int {
    for i in 0..<alphabet.count {
        if alphabet[i] == letter {
            return i
        }
    }
}

alphabet is just an array that is the alphabet in a series of strings.

Any ideas of why it is doing this? Thanks!

Answer 1

The problem lies on this line:

output = output.append(alphabet[seekAlphabet(letter: input[i])])

The append instance method mutates the original string, it does not return a value with the result. If you would have looked over the docs , you would have seen that it is declared as:

mutating func append(_ other: String)

---

Here you have two ways to solve this, by replacing

output = output.append(alphabet[seekAlphabet(letter: input[i])])

with either of the following:

• output.append(alphabet[seekAlphabet(letter: input[i])])
• output = output + alphabet[seekAlphabet(letter: input[i])]

---

Why was that assignment illegal?

Now, you know that append(_:) is a mutating function, and

output.append(alphabet[seekAlphabet(letter: input[i])])

is a () (a Void ), because it just mutates and does not return anything. Therefore, by trying to assign output to the append(_:) method, you try to assign a string to a Void , which is illegal.

Answer 2

Swift String class append(_:) is mutating function and doesn't return anything. Change this line

output = output.append(alphabet[seekAlphabet(letter: input[i])])

to

output.append(alphabet[seekAlphabet(letter: input[i])])